In the circuit shown in the figure, $C = 6\,\mu F$. The charge stored in the capacitor of capacity $C$ is..........$\mu C$
Medium
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$0-\frac{Q}{C}+10-\frac{Q}{2 C}=0$
$Q=\frac{20 C}{3}=\frac{20 \times 6}{3}=40 \mu C$
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